Question Number 56708 by gunawan last updated on 22/Mar/19 $$\mathrm{for}\:\mathrm{all}\:{n}\:\in\:\mathbb{N},\:{f}_{{n}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\mathrm{for}\:\mathrm{any}\:−\mathrm{1}<{x}<\mathrm{1}.\:\mathrm{If}\:{f}\::\:\left(−\mathrm{1},\mathrm{1}\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{with}\:{f}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}_{{n}\:} \mathrm{of}\:\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right)\:{dx}=… \\…
Question Number 56707 by gunawan last updated on 22/Mar/19 $${f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\:\:\mathrm{for}\:\mathrm{all}\:{x}\:\in\:\mathbb{R} \\ $$$${x}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{6}}\:\mathrm{for}\:\mathrm{all}\:{n}\:\in\mathbb{N} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({x}_{{n}} \right)=… \\ $$ Answered by…
Question Number 56706 by gunawan last updated on 22/Mar/19 $${f}\::\:\left[\mathrm{0},\:\mathrm{1}\right)\:\rightarrow\:\mathbb{R} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\frac{{x}}{\mathrm{2}}\right)}{{x}}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{x}}\:=… \\ $$ Commented by Abdo msup.…
Question Number 122236 by benjo_mathlover last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{mx}\right)^{{n}} −\left(\mathrm{1}+{nx}\right)^{{m}} }{{x}^{\mathrm{2}} }\:? \\ $$ Answered by Dwaipayan Shikari last updated on 15/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 122241 by TANMAY PANACEA last updated on 15/Nov/20 $${limit}… \\ $$ Commented by TANMAY PANACEA last updated on 15/Nov/20 Commented by Dwaipayan Shikari…
Question Number 122235 by benjo_mathlover last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{{x}} ?\: \\ $$ Answered by liberty last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2}}\right)^{\mathrm{x}} =\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2}}\:−\mathrm{1}\right).\mathrm{x}}…
Question Number 122192 by physicstutes last updated on 14/Nov/20 $$\mathrm{Prove}\:\mathrm{using}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{law}\:\mathrm{of}\:\mathrm{functions}\:\mathrm{that}\: \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\:. \\ $$$$\:\mathrm{Recall}\:\mathrm{that}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{theorem}\:\mathrm{states}\:\mathrm{that}\:\mathrm{if} \\ $$$$\:{f}\left({x}\right)\:\leqslant\:\mathrm{g}\left({x}\right)\:\leqslant\:{h}\left({x}\right)\: \\ $$$$\:\mathrm{and}\:\underset{{x}−{a}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{h}\left({x}\right)\:=\:{L} \\ $$$$\mathrm{then}\:,\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:{L}. \\…
Question Number 122152 by liberty last updated on 14/Nov/20 $$\:\left(\mathrm{1}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{3}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}}}\:\right)\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}\:\sqrt[{\mathrm{n}}]{\mathrm{n}}\:−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:=? \\ $$ Answered by benjo_mathlover…
Question Number 122130 by bemath last updated on 14/Nov/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{3}} ? \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)}{\left({x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:? \\ $$ Answered by TANMAY…
Question Number 187645 by Mingma last updated on 19/Feb/23 Answered by Ar Brandon last updated on 19/Feb/23 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{cos}{x}\right)}{\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}{x}\right)\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\frac{{x}^{\mathrm{8}}…